A person observes the angle of elevation of the peak of a hill from a station to be - He walks c metres along a slope inclined at an angle - and finds the angle of elevation of the peak of the hill to be - Show that the height of the peak above the ground is c sin-sin-sin-

Suppose, AB is a peak whose height above the ground is t+x. In nbsp;△DFC,sinβ=xc nbsp;⇒x=csinβand nbsp;tanβ=xy⇒y=xtanβ=csinβsinβ×cosβ=ccosβ nbsp; nb ...

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Standard X

Mathematics

Horizontal Level and Line of Sight

A person obse...

Question

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is $\frac{c \sin α \sin (γ - β)}{(\sin γ - α)}$ .

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Solution

Suppose, AB is a peak whose height above the ground is t+x.

$In △ D F C, \sin β = \frac{x}{c} \Rightarrow x = c \sin β and \tan β = \frac{x}{y} \Rightarrow y = \frac{x}{\tan β} = \frac{c \sin β}{\sin β} \times \cos β = c \cos β . . . (1) In ∆ A D E, \tan γ = \frac{t}{z} \Rightarrow z = t \cot γ . . . (2) In ∆ A B C, \tan α = \frac{t + x}{y + z} \Rightarrow t + x = (c \cos β \tan α + t c o t γ) \tan α (from (1) a n d (2)) \Rightarrow t - t c o t γ \tan α = c \cos β \tan α - c \sin β (∵ x = c \sin β) \Rightarrow t (1 - \frac{\sin α \cos γ}{\cos α \sin γ}) = c (\frac{\cos β \sin α - \cos α \sin β}{\cos α}) \Rightarrow t (\frac{\sin γ \cos α - \sin α \cos γ}{\cos α \sin γ}) = c \frac{\sin (α - β)}{\cos α} \Rightarrow t \frac{\sin (γ - β)}{\cos α \sin γ} = c \frac{\sin (α - β)}{\cos α} \Rightarrow t = c \frac{\sin γ \sin (α - β)}{\sin (γ - β)} . . . (3) Now, A B = t + x = c \frac{\sin γ \sin (α - β)}{\sin (γ - β)} + c \sin β (using (3)) = c (\frac{\sin γ \sin (α - β)}{\sin (γ - β)} + \sin β) = c [\frac{\sin γ \sin (α - β) + \sin β \sin (γ - β)}{\sin (γ - β)}] = c [\frac{\sin γ \sin α \cos β - \sin β \sin γ \cos α + \sin β \sin γ \cos α - \sin β \cos γ \sin α}{\sin (γ - β)}] = c [\frac{\sin γ \sin α \cos β - \sin β \cos γ \sin α}{\sin (γ - β)}] = c [\frac{\sin α \sin (γ - β)}{\sin (γ - β)}] = \frac{c \sin α \sin (γ - β)}{\sin (γ - β)} Hence proved .$

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A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is c sin α sin γ-βsin γ-α.